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Tentukan nilia integral dari

Pertanyaan

Tentukan nilia integral dari integral fraction numerator 3 x plus 2 over denominator square root of x end fraction d x space equals..

Pembahasan Soal:

integral fraction numerator 3 x plus 2 over denominator square root of x end fraction d x space equals integral 3 x plus 2 space d x plus integral x to the power of fraction numerator negative 1 over denominator 2 end fraction end exponent d x equals

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Rahmawati

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Tentukan turunan pertama dari fungsi berikut.

Pembahasan Soal:

Dengan menggunakan rumus turunan fungsi, maka berlaku:

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell u over v end cell row cell f apostrophe left parenthesis x right parenthesis end cell equals cell fraction numerator u apostrophe v minus u v apostrophe over denominator v squared end fraction end cell end table 

Misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell 3 x squared minus 5 end cell row cell u apostrophe end cell equals cell 6 x end cell row blank blank blank row v equals cell x plus 6 end cell row cell v apostrophe end cell equals 1 end table 

Sehingga, turunan pertama dari fungsi tersebut adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell f apostrophe left parenthesis x right parenthesis end cell equals cell fraction numerator u apostrophe v minus u v apostrophe over denominator v squared end fraction end cell row blank equals cell fraction numerator 6 x left parenthesis x plus 6 right parenthesis minus left parenthesis 3 x squared minus 5 right parenthesis 1 over denominator left parenthesis x plus 6 right parenthesis squared end fraction end cell row blank equals cell fraction numerator 6 x squared plus 36 x minus 3 x squared plus 5 over denominator x squared plus 12 x plus 36 end fraction end cell row cell f apostrophe left parenthesis x right parenthesis end cell equals cell fraction numerator 3 x squared plus 36 x plus 5 over denominator x squared plus 12 x plus 36 end fraction end cell end table 

Dengan demikian, turunan pertama dari fungsi pada soal tersebut adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 x squared plus 36 x plus 5 over denominator x squared plus 12 x plus 36 end fraction end cell end table. 

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Pembahasan Soal:

Untuk menentukan turunan dari bentuk di atas, maka kita dapat menyederhanakan bentuk tersebut:

table attributes columnalign right center left columnspacing 0px end attributes row cell integral open parentheses fraction numerator x to the power of 4 minus 1 over denominator x cubed plus x end fraction close parentheses squared d x end cell equals cell integral open parentheses fraction numerator x to the power of 4 minus 1 over denominator x open parentheses x squared plus 1 close parentheses end fraction close parentheses squared d x end cell row blank equals cell integral open parentheses fraction numerator open parentheses x squared minus 1 close parentheses open parentheses x squared plus 1 close parentheses over denominator x open parentheses x squared plus 1 close parentheses end fraction close parentheses squared d x end cell row blank equals cell integral open parentheses fraction numerator open parentheses x squared minus 1 close parentheses over denominator x end fraction close parentheses squared d x end cell row blank equals cell integral open parentheses x squared over x minus 1 over x close parentheses squared d x end cell row blank equals cell integral open parentheses x minus 1 over x close parentheses squared d x end cell row blank equals cell integral x squared minus 2 x times 1 over x plus open parentheses 1 over x close parentheses squared space d x end cell row blank equals cell integral x squared minus 2 plus x to the power of negative 2 end exponent space d x end cell row blank equals cell 1 third x cubed minus 2 x minus x to the power of negative 1 end exponent plus c end cell row blank equals cell 1 third x cubed minus 2 x minus 1 over x plus c end cell row blank equals blank end table

Jadi, hasil dari Error converting from MathML to accessible text. adalah 1 third x cubed minus 2 x minus 1 over x plus c

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Pembahasan Soal:

Bentuk integral di atas dapat diselesaikan seperti berikut,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell integral fraction numerator 2 x cubed minus 3 x squared plus 1 over denominator x squared end fraction space d x end cell row blank equals cell integral fraction numerator 2 x cubed over denominator x squared end fraction minus fraction numerator 3 x squared over denominator x squared end fraction plus 1 over x squared d x end cell row blank equals cell integral 2 x minus 3 plus x to the power of negative 2 end exponent space d x end cell row blank equals cell 2 over 2 x squared minus 3 x plus left parenthesis negative 1 right parenthesis x to the power of negative 1 end exponent plus C end cell row blank equals cell x squared minus 3 x minus 1 over x plus C end cell end table end style 

Dengan demikian, hasil dari integral tersebut adalah begin mathsize 14px style x squared minus 3 x minus 1 over x plus C end style.

Jadi, jawaban yang tepat adalah B.

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Hasil dari

Pembahasan Soal:

Dengan menggunakan konsep integral tak tentu, maka misalkan:

table attributes columnalign right center left columnspacing 0px end attributes row u equals cell 9 x squared minus 6 x plus 12 end cell row cell d u end cell equals cell 18 x minus 6 space end cell end table

Selanjutnya,

6 x minus 2 d x equals 1 third d u

Jadi,

table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator left parenthesis 6 x minus 2 right parenthesis over denominator left parenthesis 9 x squared minus 6 x plus 12 right parenthesis cubed end fraction d x end cell equals cell integral left parenthesis 6 x minus 2 right parenthesis left parenthesis 9 x squared minus 6 x plus 12 right parenthesis to the power of negative 3 end exponent d x end cell row blank equals cell integral 1 third u to the power of negative 3 end exponent d u end cell row blank equals cell fraction numerator 1 over denominator 3 cross times left parenthesis negative 2 right parenthesis end fraction u to the power of negative 2 end exponent plus c end cell row blank equals cell fraction numerator 1 over denominator negative 6 end fraction u to the power of negative 2 end exponent plus c end cell row blank equals cell fraction numerator 1 over denominator negative 6 end fraction open parentheses 9 x squared minus 6 x plus 12 close parentheses to the power of negative 2 end exponent plus c end cell end table

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Hasil dari  adalah ....

Pembahasan Soal:

Perhatikan perhitungan berikut.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell integral fraction numerator x to the power of 4 plus 3 x cubed minus 8 x squared minus 12 x plus 16 over denominator x squared minus 4 end fraction blank d x end cell equals cell integral fraction numerator open parentheses x plus 2 close parentheses open parentheses x minus 2 close parentheses open parentheses x squared plus 3 x minus 4 close parentheses over denominator open parentheses x minus 2 close parentheses open parentheses x plus 2 close parentheses end fraction blank d x end cell row blank equals cell integral open parentheses x squared plus 3 x minus 4 close parentheses blank d x end cell row blank equals cell integral x squared blank d x plus integral 3 x blank d x minus integral 4 blank d x end cell row blank equals cell open parentheses fraction numerator 1 over denominator 2 plus 1 end fraction x to the power of 2 plus 1 end exponent plus C subscript 1 close parentheses plus 3 open parentheses fraction numerator 1 over denominator 1 plus 1 end fraction x to the power of 1 plus 1 end exponent plus C subscript 2 close parentheses minus open parentheses 4 x plus C subscript 3 close parentheses end cell row blank equals cell x cubed over 3 plus 3 over 2 x squared minus 4 x plus C subscript 1 plus 3 C subscript 2 minus C subscript 3 end cell end table end style 

Misalkan begin mathsize 14px style C subscript 1 plus 3 C subscript 2 minus C subscript 3 equals C comma end style maka

begin mathsize 14px style x cubed over 3 plus 3 over 2 x squared minus 4 x plus C subscript 1 plus 3 C subscript 2 minus C subscript 3 equals x cubed over 3 plus 3 over 2 x squared minus 4 x plus C end style 

Jadi, jawaban yang tepat adalah D.

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